ta có \(45.10^{-9}=q_1+q_2\)

\(F=4,5.10^{-3}=k.\dfrac{\left|q_1q_2\right|}{0,06^2}\)

từ 2 pt trên \(\Rightarrow\left\{{}\begin{matrix}q_1=2,23.10^{-8}\\q_2=-6,73.10^{-8}\end{matrix}\right.\)

25,D

26,A

\(\overrightarrow{F_{12}}=k.\dfrac{q_1q_2}{0,1^2}=...\)

\(F_1=k.\dfrac{q^2}{R^2}\Rightarrow q=2,6.10^{-9}\)

\(\Rightarrow F_2=2,5.10^{-4}=k.\dfrac{q^2}{x^2}\Rightarrow x=...\)

ta có \(h=\dfrac{1}{2}gt^2=125\left(m\right)\)

\(v_t=gt=50\left(m/s\right)\)

\(\dfrac{50}{5+20+R_3}=1=>R_3=25\)

\(U_1=1.5=5\left(V\right);U_2=1.20=20\left(V\right);U_3=1.25=25\left(V\right)\)

c, ABC là tam giác cân tại C

\(F=2.k.\dfrac{\left|q_1q_3\right|}{0,015^2}.\dfrac{1,5}{1,5}=12,8\left(N\right)\)

a, ta thấy CA+CB=AB

\(F_1+F_2=F=k.\left(\dfrac{\left|q_1q_3\right|}{CA^2}+\dfrac{\left|q_2q_3\right|}{CB^2}\right)=14,4+3,6=18\left(N\right)\)

b, CA+AB=CB

\(F=F_1-F_2=k.\left(\dfrac{\left|q_1q_3\right|}{CA^2}-\dfrac{\left|q_2q_3\right|}{CB^2}\right)=3,6-0,567=...\left(N\right)\)

 pt cân bằng cho 3 lần đổ

\(10mC.\left(t-t_k-8\right)=m_kC_k8\left(1\right)\)

\(2mC.\left(t-t_k-3\right)=m_kC_k.3\left(2\right)\)

\(mC.\left(t-t_k-t_x\right)=m_kC_k.t_x\left(3\right)\)

chia 1 cho 2 

\(t-t_k=28\)

chia 3 cho 2 

\(\dfrac{28-t_x}{52}=\dfrac{t_x}{3}\Rightarrow t_x=...\)

Khánh Nam.....! ( IDΣΛ...                                                          hỏi câu nào sai mà :))